Question: Find $ \lim_{x\to\infty} \left(3x+1\right)^{^{\frac4x}}$. Choose 1 answer: Choose 1 answer: (Choice A) A $1$ (Choice B) B $\dfrac43$ (Choice C) C $\dfrac34$ (Choice D) D The limit doesn't exist.
Explanation: Taking $x$ to $\infty$ in $(3x+1)^{^{\frac{4}{x}}}$ results in the indeterminate form $\infty^{^{0}}$. To make the expression easier to analyze, let's take its natural log (this is a common trick when dealing with composite exponential functions). In other words, letting $y=(3x+1)^{^{\frac{4}{x}}}$, we will find $\lim_{x\to \infty}\ln(y)$. Once we find it, we will be able to find $\lim_{x\to \infty}y$. $\ln(y) =\dfrac{4\ln(3x+1)}{x}$ Taking $x$ to $\infty$ in $\dfrac{4\ln(3x+1)}{x}$ results in the indeterminate form $\dfrac{\infty}{\infty}$, so now it's L'Hôpital's rule's turn to help us with our quest! $\begin{aligned} &\phantom{=}\lim_{x\to \infty}\ln(y) \\\\ &=\lim_{x\to \infty}\dfrac{4\ln(3x+1)}{x} \\\\ &=\lim_{x\to \infty}\dfrac{\dfrac{d}{dx}[4\ln(3x+1)]}{\dfrac{d}{dx}[x]} \gray{\text{L'Hôpital's rule}} \\\\ &=\lim_{x\to \infty}\dfrac{\left(\dfrac{12}{3x+1}\right)}{1} \\\\ &=0 \end{aligned}$ Note that we were only able to use L'Hôpital's rule because the limit $\lim_{x\to \infty}\dfrac{\dfrac{d}{dx}[4\ln(3x+1)]}{\dfrac{d}{dx}[x]}$ actually exists. We found that $\lim_{x\to \infty}\ln(y)=0$, which means $\lim_{x\to \infty}y=1$. [Why?] In conclusion, $\lim_{x\to \infty}\dfrac{4\ln(3x+1)}{x}=1$.